Part 8: Hypothesis Testing 8-1/50 Econometrics I Professor William Greene Stern School of Business Department of Economics

Part 8: Hypothesis Testing8-1/50

Econometrics IProfessor William Greene

Stern School of Business

Department of Economics


Econometrics I

Part 8 – Interval Estimation and Hypothesis Testing


Interval Estimation

b = point estimator of We acknowledge the sampling variability.

Estimated sampling variance b = + sampling variability induced by


Point estimate is only the best single guess Form an interval, or range of plausible values Plausible likely values with acceptable degree

of probability. To assign probabilities, we require a distribution

for the variation of the estimator. The role of the normality assumption for


Confidence Intervalbk = the point estimate

Std.Err[bk] = sqr{[σ2(XX)-1]kk} = vk

Assume normality of ε for now: bk ~ N[βk,vk

2] for the true βk. (bk-βk)/vk ~ N[0,1]

Consider a range of plausible values of βk given the point estimate bk. bk sampling error. Measured in standard error units, |(bk – βk)/ vk| < z* Larger z* greater probability (“confidence”) Given normality, e.g., z* = 1.96 95%, z*=1.64590% Plausible range for βk then is bk ± z* vk


Computing the Confidence IntervalAssume normality of ε for now:

bk ~ N[βk,vk2] for the true βk.

(bk-βk)/vk ~ N[0,1]

vk = [σ2(X’X)-1]kk is not known because σ2 must be estimated.

Using s2 instead of σ2, (bk-βk)/Est.(vk) ~ t[N-K].

(Proof: ratio of normal to sqr(chi-squared)/df is pursued in your text.)

Use critical values from t distribution instead of standard normal. Will be the same as normal if N > 100.


Confidence Interval

Critical t[.975,29] = 2.045Confidence interval based on t: 1.27365 2.045 * .1501Confidence interval based on normal: 1.27365 1.960 * .1501


Bootstrap Confidence Interval For a Coefficient


Bootstrap CI for Least Squares


Bootstrap CI for Least Absolute Deviations


Testing a Hypothesis Using a Confidence Interval

Given the range of plausible values

Testing the hypothesis that a coefficient equals zero or some other particular value:

Is the hypothesized value in the confidence interval?

Is the hypothesized value within the range of plausible values? If not, reject the hypothesis.


Test a Hypothesis About a Coefficient


Classical Hypothesis Testing

We are interested in using the linear regression to support or cast doubt on the validity of a theory about the real world counterpart to our statistical model. The model is used to test hypotheses about the underlying data generating process.


Types of Tests

Nested Models: Restriction on the parameters of a particular modely = 1 + 2x + 3z + , 3 = 0

Nonnested models: E.g., different RHS variablesyt = 1 + 2xt + 3xt-1 + t

yt = 1 + 2xt + 3yt-1 + wt Specification tests:

~ N[0,2] vs. some other distribution


Methodology

Bayesian Prior odds compares strength of prior beliefs in two states of the

world Posterior odds compares revised beliefs Symmetrical treatment of competing ideas Not generally practical to carry out in meaningful situations

Classical “Null” hypothesis given prominence Propose to “reject” toward default favor of “alternative” Asymmetric treatment of null and alternative Huge gain in practical applicability


Neyman – Pearson Methodology

Formulate null and alternative hypotheses Define “Rejection” region = sample evidence

that will lead to rejection of the null hypothesis. Gather evidence Assess whether evidence falls in rejection region

or not.


Inference in the Linear Model

Formulating hypotheses: linear restrictions as a general framework

Hypothesis Testing J linear restrictions

Analytical framework: y = X + Hypothesis: R - q = 0,

Substantive restrictions: What is a "testable hypothesis?" Substantive restriction on parameters Reduces dimension of parameter space Imposition of restriction degrades estimation criterion


Testable Implications of a Theory Investors care about interest rates and expected

inflation:I = b1 + b2r + b3dp + e

Investors care about real interest ratesI = c1 + c2(r-dp) + c3dp + e

No testable restrictions implied. c1 =b1, c2=b2-b3, c3=b3.

Investors care only about real interest ratesI = f1 + f2(r-dp) + f3dp + e. f3 = 0


The General Linear Hypothesis: H0: R - q = 0

A unifying departure point: Regardless of the hypothesis, least squares is unbiased.

E[b] =

The hypothesis makes a claim about the population

R – q = 0. Then, if the hypothesis is true, E[Rb – q] = 0.

The sample statistic, Rb – q will not equal zero.Two possibilities:

Rb – q is small enough to attribute to sampling variabilityRb – q is too large (by some measure) to be plausibly attributed to

sampling variabilityLarge Rb – q is the rejection region.


Approaches to Defining the Rejection Region

(1) Imposing the restrictions leads to a loss of fit. R2 must go down. Does it go down “a lot?” (I.e., significantly?).

Ru2 = unrestricted model, Rr

2 = restricted model fit. F = { (Ru

2 – Rr2)/J } / [(1 – Ru

2)/(N-K)] = F[J,N-K].

(2) Is Rb - q close to 0? Basing the test on the discrepancy vector: m = Rb - q. Using the Wald criterion: m(Var[m])-1m

has a chi-squared distribution with J degrees of freedom But, Var[m] = R[2(X’X)-1]R. If we use our estimate of 2, we get an F[J,N-K], instead. (Note, this is based on using ee/(N-K) to estimate 2.)

These are the same for the linear model


Testing Fundamentals - I SIZE of a test = Probability it will incorrectly

reject a “true” null hypothesis. This is the probability of a Type I error.

Under the null hypothesis, F(3,100) has an F distribution with (3,100) degrees of freedom. Even if the null is true, F will be larger than the 5% critical value of 2.7 about 5% of the time.


Testing ProceduresHow to determine if the statistic is 'large.' Need a 'null distribution.'If the hypothesis is true, then the statistic will have a certain

distribution. This tells you how likely certain values are, and in particular, if the hypothesis is true, then 'large values' will be unlikely.

If the observed statistic is too large, conclude that the assumed distribution must be incorrect and the hypothesis should be rejected.

For the linear regression model with normally distributed disturbances, the distribution of the relevant statistic is F with J and N-K degrees of freedom.


Distribution Under the Null

Density of F[3,100]

X

.250

.500

.750

.000

1 2 3 40

FDENSIT

Y


A Simulation Experiment

sample ; 1 - 100 $matrix ; fvalues=init(1000,1,0)$proc$create ; fakelogc = rnn(-.319557,1.54236)$ Coefficients all = 0regress ; quietly ; lhs = fakelogc Compute regression ; rhs=one,logq,logpl_pf,logpk_pf$calc ; fstat = (rsqrd/3)/((1-rsqrd)/(n-4))$ Compute Fmatrix ; fvalues(i)=fstat$ Save 1000 Fsendprocexecute ; i= 1,1000 $ 1000 replicationshistogram ; rhs = fvalues ; title=F Statistic for H0:b2=b3=b4=0$


Simulation Results

48 outcomes to the right of 2.7 in this run of the experiment.


Testing Fundamentals - II

POWER of a test = the probability that it will correctly reject a “false null” hypothesis

This is 1 – the probability of a Type II error. The power of a test depends on the specific

alternative.


Power of a Test

Null: Mean = 0. Reject if observed mean < -1.96 or > +1.96.Prob(Reject null|mean=0) = 0.05

Prob(Reject null|mean=.5)=0.07902

Prob(Reject null|mean=1)=0.170066. Increases as the (alternative) mean rises.

B E TA

.084

.168

.251

.335

.419

.000

-3 -2 -1 0 1 2 3 4 5-4

N2 N0 N1

Variable


Test Statistic

For the fit measures, use a normalized measure of the loss of fit:

r rr

r rr r

0 since

and

0 since

2 2u r 2 2

u r2u

2 2u uu

yy yy

u uu u

u u

R -R / JF[J,n -K] = R R

1-R / (N-K)

Often useful

R =1- R =1-S S

Insert these in F and it becomes

/ JF[J,n -K] =

/ (N-K)

e e e e

e e e ee e e e

e e


Test StatisticsForming test statistics:

For distance measures use Wald type of distance measure, W = m[Est.Var(m)]-1m

An important relationship between t and F

For a single restriction, m = r’b - q. The variance is r’(Var[b])r

The distance measure is m / standard error of m.


An important relationship between t and F

2

Chi Squared[J] / JF

Chi squared[N K] / (N K)

where the two chi-squared variables are independent.

If J = 1, i.e., testing a single restriction,

Chi Squared[1] /1F


(N[0,1])

Chi s

2

2

quared[N K] / (N K)

N[0,1] = t[1]


For a single restriction, F[1,N-K] is the square of the t ratio.


ApplicationTime series regression,

LogG = 1 + 2logY + 3logPG

+ 4logPNC + 5logPUC + 6logPPT

+ 7logPN + 8logPD + 9logPS +

Period = 1960 - 1995. Note that all coefficients in the model are elasticities.


Full Model----------------------------------------------------------------------Ordinary least squares regression ............LHS=LG Mean = 5.39299 Standard deviation = .24878 Number of observs. = 36Model size Parameters = 9 Degrees of freedom = 27Residuals Sum of squares = .00855 <******* Standard error of e = .01780 <*******Fit R-squared = .99605 <******* Adjusted R-squared = .99488 <*******--------+-------------------------------------------------------------Variable| Coefficient Standard Error t-ratio P[|T|>t] Mean of X--------+-------------------------------------------------------------Constant| -6.95326*** 1.29811 -5.356 .0000 LY| 1.35721*** .14562 9.320 .0000 9.11093 LPG| -.50579*** .06200 -8.158 .0000 .67409 LPNC| -.01654 .19957 -.083 .9346 .44320 LPUC| -.12354* .06568 -1.881 .0708 .66361 LPPT| .11571 .07859 1.472 .1525 .77208 LPN| 1.10125*** .26840 4.103 .0003 .60539 LPD| .92018*** .27018 3.406 .0021 .43343 LPS| -1.09213*** .30812 -3.544 .0015 .68105--------+-------------------------------------------------------------


Test About One Parameter

Is the price of public transportation really relevant? H0 : 6 = 0.

Confidence interval: b6 t(.95,27) Standard error = .11571 2.052(.07859) = .11571 .16127 = (-.045557 ,.27698)

Contains 0.0. Do not reject hypothesis

Regression fit if drop? Without LPPT, R-squared= .99573 Compare R2, was .99605,

F(1,27) = [(.99605 - .99573)/1]/[(1-.99605)/(36-9)] = 2.187 = 1.4722 (with some rounding difference)


Hypothesis Test: Sum of Coefficients = 1?----------------------------------------------------------------------Ordinary least squares regression ............LHS=LG Mean = 5.39299 Standard deviation = .24878 Number of observs. = 36Model size Parameters = 9 Degrees of freedom = 27Residuals Sum of squares = .00855 <******* Standard error of e = .01780 <*******Fit R-squared = .99605 <******* Adjusted R-squared = .99488 <*******--------+-------------------------------------------------------------Variable| Coefficient Standard Error t-ratio P[|T|>t] Mean of X--------+-------------------------------------------------------------Constant| -6.95326*** 1.29811 -5.356 .0000 LY| 1.35721*** .14562 9.320 .0000 9.11093 LPG| -.50579*** .06200 -8.158 .0000 .67409 LPNC| -.01654 .19957 -.083 .9346 .44320 LPUC| -.12354* .06568 -1.881 .0708 .66361 LPPT| .11571 .07859 1.472 .1525 .77208 LPN| 1.10125*** .26840 4.103 .0003 .60539 LPD| .92018*** .27018 3.406 .0021 .43343 LPS| -1.09213*** .30812 -3.544 .0015 .68105--------+-------------------------------------------------------------


Imposing the Restriction----------------------------------------------------------------------Linearly restricted regressionLHS=LG Mean = 5.392989 Standard deviation = .2487794 Number of observs. = 36Model size Parameters = 8 <*** 9 – 1 restriction Degrees of freedom = 28Residuals Sum of squares = .0112599 <*** With the restrictionResiduals Sum of squares = .0085531 <*** Without the restrictionFit R-squared = .9948020Restrictns. F[ 1, 27] (prob) = 8.5(.01)Not using OLS or no constant.R2 & F may be < 0--------+-------------------------------------------------------------Variable| Coefficient Standard Error t-ratio P[|T|>t] Mean of X--------+-------------------------------------------------------------Constant| -10.1507*** .78756 -12.889 .0000 LY| 1.71582*** .08839 19.412 .0000 9.11093 LPG| -.45826*** .06741 -6.798 .0000 .67409 LPNC| .46945*** .12439 3.774 .0008 .44320 LPUC| -.01566 .06122 -.256 .8000 .66361 LPPT| .24223*** .07391 3.277 .0029 .77208 LPN| 1.39620*** .28022 4.983 .0000 .60539 LPD| .23885 .15395 1.551 .1324 .43343 LPS| -1.63505*** .27700 -5.903 .0000 .68105--------+-------------------------------------------------------------F = [(.0112599 - .0085531)/1] / [.0085531/(36 – 9)] = 8.544691


Joint Hypotheses

Joint hypothesis: Income elasticity = +1, Own price elasticity = -1.The hypothesis implies that logG = β1 + logY – logPg + β4 logPNC + ...Strategy: Regress logG – logY + logPg on the other variables and

Compare the sums of squaresWith two restrictions imposedResiduals Sum of squares = .0286877Fit R-squared = .9979006UnrestrictedResiduals Sum of squares = .0085531Fit R-squared = .9960515

F = ((.0286877 - .0085531)/2) / (.0085531/(36-9)) = 31.779951The critical F for 95% with 2,27 degrees of freedom is 3.354. The hypothesis is rejected.


Basing the Test on R2

After building the restrictions into the model and computing restricted and unrestricted regressions: Based on R2s,

F = ((.9960515 - .997096)/2)/((1-.9960515)/(36-9)) = -3.571166 (!)

What's wrong? The unrestricted model used LHS = logG. The restricted one used logG-logY. The calculation is safe using the sums of squared residuals.


Wald Distance MeasureTesting more generally about a single parameter.

Sample estimate is bk

Hypothesized value is βk

How far is βk from bk? If too far, the hypothesis is inconsistent with the sample evidence.

Measure distance in standard error units

t = (bk - βk)/Estimated vk.If t is “large” (larger than critical value), reject the

hypothesis.


The Wald Statistic

-1

Most test statistics are Wald distance measures

W = (random vector - hypothesized value)' times

[Variance of difference] times

(random vector - hypothesized value)

0 0 -1 0

= Normalized distance measure

= ( - ) [Var( - )] ( - )

Distributed as chi-squared(J ) if (1) the distance is

normally distributed and (2) the variance matrix is

the true one, not the esti

q q ' q q q q

mate.


Robust Tests The Wald test generally will (when properly

constructed) be more robust to failures of the narrow model assumptions than the t or F

Reason: Based on “robust” variance estimators and asymptotic results that hold in a wide range of circumstances.

Analysis: Later in the course – after developing asymptotics.


Particular CasesSome particular cases:One coefficient equals a particular value: F = [(b - value) / Standard error of b ]2 = square of familiar t ratio.

Relationship is F [ 1, d.f.] = t2[d.f.]A linear function of coefficients equals a particular value

(linear function of coefficients - value)2

F = ---------------------------------------------------- Variance of linear function

Note square of distance in numeratorSuppose linear function is k wk bk

Variance is kl wkwl Cov[bk,bl]This is the Wald statistic. Also the square of the somewhat

familiar t statistic.Several linear functions. Use Wald or F. Loss of fit measures may be easier to

compute.


Hypothesis Test: Sum of CoefficientsDo the three aggregate price elasticities sum to zero?H0 :β7 + β8 + β9 = 0R = [0, 0, 0, 0, 0, 0, 1, 1, 1], q = [0]

Variable| Coefficient Standard Error t-ratio P[|T|>t] Mean of X--------+------------------------------------------------------------- LPN| 1.10125*** .26840 4.103 .0003 .60539 LPD| .92018*** .27018 3.406 .0021 .43343 LPS| -1.09213*** .30812 -3.544 .0015 .68105


Wald Test


Using the Wald Statistic--> Matrix ; R = [0,1,0,0,0,0,0,0,0 / 0,0,1,0,0,0,0,0,0]$--> Matrix ; q = [1/-1]$--> Matrix ; list ; m = R*b - q $Matrix M has 2 rows and 1 columns. 1 +-------------+ 1| .35721 2| .49421 +-------------+--> Matrix ; list ; vm = R*varb*R' $Matrix VM has 2 rows and 2 columns. 1 2 +-------------+-------------+ 1| .02120 .00291 2| .00291 .00384 +-------------+-------------+--> Matrix ; list ; w = m'<vm>m $Matrix W has 1 rows and 1 columns. 1 +-------------+ 1| 63.55962 +-------------+

Joint hypothesis:b(LY) = 1b(LPG) = -1


Application: Cost Function


Regression Results----------------------------------------------------------------------Ordinary least squares regression ............LHS=C Mean = 3.07162 Standard deviation = 1.54273 Number of observs. = 158Model size Parameters = 9 Degrees of freedom = 149Residuals Sum of squares = 2.56313 Standard error of e = .13116Fit R-squared = .99314 Adjusted R-squared = .99277--------+-------------------------------------------------------------Variable| Coefficient Standard Error t-ratio P[|T|>t] Mean of X--------+-------------------------------------------------------------Constant| 5.22853 3.18024 1.644 .1023 K| -.21055 .21757 -.968 .3348 4.25096 L| -.85353*** .31485 -2.711 .0075 8.97280 F| .18862 .20225 .933 .3525 3.39118 Q| -.96450*** .36798 -2.621 .0097 8.26549 Q2| .05250*** .00459 11.430 .0000 35.7913 QK| .04273 .02758 1.550 .1233 35.1677 QL| .11698*** .03728 3.138 .0021 74.2063 QF| .05950** .02478 2.401 .0176 28.0108--------+-------------------------------------------------------------Note: ***, **, * = Significance at 1%, 5%, 10% level.----------------------------------------------------------------------


Price Homogeneity: Only Price Ratios Matter β2 + β3 + β4 = 1. β7 + β8 + β9 = 0.----------------------------------------------------------------------Linearly restricted regression....................LHS=C Mean = 3.07162 Standard deviation = 1.54273 Number of observs. = 158Model size Parameters = 7 Degrees of freedom = 151Residuals Sum of squares = 2.85625 Standard error of e = .13753Fit R-squared = .99236Restrictns. F[ 2, 149] (prob) = 8.5(.0003)Not using OLS or no constant. Rsqrd & F may be < 0--------+-------------------------------------------------------------Variable| Coefficient Standard Error t-ratio P[|T|>t] Mean of X--------+-------------------------------------------------------------Constant| -7.24078*** 1.01411 -7.140 .0000 K| .38943** .16933 2.300 .0228 4.25096 L| .19130 .19530 .979 .3289 8.97280 F| .41927** .20077 2.088 .0385 3.39118 Q| .45889*** .12402 3.700 .0003 8.26549 Q2| .06008*** .00441 13.612 .0000 35.7913 QK| -.02954 .02125 -1.390 .1665 35.1677 QL| -.00462 .02364 -.195 .8454 74.2063 QF| .03416 .02489 1.373 .1720 28.0108--------+-------------------------------------------------------------


Imposing the RestrictionsAlternatively, compute the restricted regression by converting to price ratios andImposing the restrictions directly. This is a regression of log(c/pf) on log(pk/pf), log(pl/pf) etc.----------------------------------------------------------------------Ordinary least squares regression ............LHS=LOGC_PF Mean = -.31956 Standard deviation = 1.54236 Number of observs. = 158Model size Parameters = 7 Degrees of freedom = 151Residuals Sum of squares = 2.85625 (restricted)Residuals Sum of squares = 2.56313 (unrestricted)--------+-------------------------------------------------------------Variable| Coefficient Standard Error t-ratio P[|T|>t] Mean of X--------+-------------------------------------------------------------Constant| -7.24078*** 1.01411 -7.140 .0000 LOGQ| .45889*** .12402 3.700 .0003 8.26549 LOGQSQ| .06008*** .00441 13.612 .0000 35.7913LOGPK_PF| .38943** .16933 2.300 .0228 .85979LOGPL_PF| .19130 .19530 .979 .3289 5.58162LQ_LPKPF| -.02954 .02125 -1.390 .1665 7.15696LQ_LPLPF| -.00462 .02364 -.195 .8454 46.1956--------+-------------------------------------------------------------F Statistic = ((2.85625 – 2.56313)/2) / (2.56313/(158-9)) = 8.52(This is a problem. The underlying theory requires linear homogeneity in prices.)


Wald Test of the RestrictionsChi squared = J*F

wald ; fn1 = b_k + b_l + b_f - 1 ; fn2 = b_qk + b_ql + b_qf – 0 $-----------------------------------------------------------WALD procedure. Estimates and standard errorsfor nonlinear functions and joint test ofnonlinear restrictions.Wald Statistic = 17.03978Prob. from Chi-squared[ 2] = .00020--------+--------------------------------------------------Variable| Coefficient Standard Error b/St.Er. P[|Z|>z]--------+-------------------------------------------------- Fncn(1)| -1.87546*** .45621 -4.111 .0000 Fncn(2)| .21921*** .05353 4.095 .0000--------+--------------------------------------------------Note: ***, **, * = Significance at 1%, 5%, 10% level.-----------------------------------------------------------Recall:F Statistic = ((2.85625 – 2.56313)/2) / (2.56313/(158-9)) = 8.52The chi squared statistic is JF. Here, 2F = 17.04 = the chi squared.


Test of HomotheticityCross Product Terms = 0

Homothetic Production: β7 = β8 = β9 = 0.With linearity homogeneity in prices already imposed, this is β6 = β7 = 0.

-----------------------------------------------------------WALD procedure. Estimates and standard errorsfor nonlinear functions and joint test ofnonlinear restrictions.Wald Statistic = 2.57180Prob. from Chi-squared[ 2] = .27640--------+--------------------------------------------------Variable| Coefficient Standard Error b/St.Er. P[|Z|>z]--------+-------------------------------------------------- Fncn(1)| -.02954 .02125 -1.390 .1644 Fncn(2)| -.00462 .02364 -.195 .8451--------+--------------------------------------------------

The F test would produce F = ((2.90490 – 2.85625)/1)/(2.85625/(158-7)) = 1.285

Documents

Part 8: Hypothesis Testing 8-1/50 Econometrics I Professor William Greene Stern School of Business Department of Economics